# Problem 14.5c6

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = 4k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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— This answer is my guess. —

.

The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L \end{aligned}}

The left R’+ sector has no massless states.

The right-moving NS+ states:

\displaystyle{\begin{aligned} \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\ \end{aligned}}

The R- states (that used as right-moving states):

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle \\ \end{aligned}}

~~~

Since R’+ has no massless states:

spacetime bosons:

$NS'+ \otimes NS+$

\displaystyle{\begin{aligned} \left( \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \right) \otimes \left( b_{-1/2}^I~|NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

\displaystyle{\begin{aligned} = \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} b_{-1/2}^I |NS' \rangle_L \otimes \left( |NS \rangle \otimes |p^+, \vec p_T \rangle \right) \end{aligned}}

spacetime fermions:

$NS'+ \otimes R-$

\displaystyle{\begin{aligned} \{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \} |NS' \rangle_L \otimes |R_a \rangle \\ \end{aligned}}

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— This answer is my guess. —

— Me@2018-12-18 07:46:15 PM

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2018.12.18 Tuesday (c) All rights reserved by ACHK

# Problem 14.5c5

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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We have the following well-known bosonic string mass formulae:

open string:

$\displaystyle{\alpha' M^2 = N - a}$

closed string:

$\displaystyle{\frac{1}{4} \alpha' M^2 = N_L - a}$
$\displaystyle{\frac{1}{4} \alpha' M^2 = N_R - a}$

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

.

How come there is an extra $\displaystyle{\frac{1}{4}}$ at the beginning of the closed string formula?

p.322

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

.

\displaystyle{\begin{aligned} \alpha' M_L^2 &= \alpha' M_R^2 \\ \frac{1}{2} \alpha' M^2 &= \alpha' M_L^2 + \alpha' M_R^2 \\ \alpha' M^2 &= 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2 \\ \end{aligned}}

— Me@2018-12-15 08:59:18 PM

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2018.12.15 Saturday (c) All rights reserved by ACHK

# Slope parameter

Problem 14.5c4 | Counting states in heterotic SO(32) string theory

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c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

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What is the meaning of $\displaystyle{\alpha'}$?

How come $\displaystyle{\alpha' = \frac{1}{2}}$?

— Me@2018-12-07 10:43:10 PM

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If we consider a rigidly rotating open string, $\alpha'$ is the proportionality constant that relates the angular momentum $J$ of the string, measured in units of $\hbar$, to the square of its energy $E$. More explicitly,

$\displaystyle{ \frac{J}{\hbar} = \alpha' E^2 }$

— p.68

.

$\displaystyle{ J = \frac{1}{ 2 \pi T_0 c} E^2 }$

As anticipated, the angular momentum is proportional to the square of the energy of the string. Comparing with equation (8.69) we deduce that

\displaystyle{ \begin{aligned} \alpha' &= \frac{1}{2 \pi T_0 \hbar c} \\ T_0 &= \frac{1}{2 \pi \alpha' \hbar c} \end{aligned}}

These equations relate the slope parameter $\alpha'$ to the string tension $T_0$.

— p.69

— A First Course in String Theory

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2018.12.11 Tuesday (c) All rights reserved by ACHK

# Problem 14.5c3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

.

At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$.

c) Are there tachyonic states in heterotic string theory?

Write out the massless states of the theory (bosons and fermions) …

~~~

.

Open String:

\displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ L_n^{\perp} &= \frac{1}{2} \sum_{-\infty}^{\infty} \alpha_{n-p}^I \alpha_{p}^I~~~(n \ne 0) \\ L_0^{\perp} &= \alpha' p^I p^I + N^\perp \end{aligned} }

.

Closed String:

\displaystyle{ \begin{aligned} N^\perp &= \sum_{p=1}^{\infty} \alpha_{-p}^I \alpha_p^I \\ \bar N^\perp &= \sum_{p=1}^{\infty} \bar \alpha_{-p}^I \bar \alpha_p^I \\ L_0^\perp &= \frac{\alpha'}{4} p^I p^I + N^\perp \\ \bar L_0^\perp &= \frac{\alpha'}{4} p^I p^I + \bar N^\perp \end{aligned} }

— A First Course in String Theory

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We have the following well-known bosonic string mass formulae $\displaystyle{\alpha' = \frac{1}{2}}$:

open string:

$\displaystyle{\frac{1}{2} M^2 = N - a}$

closed string:

$\displaystyle{\frac{1}{8} M^2 = N_L - a}$
$\displaystyle{\frac{1}{8} M^2 = N_R - a}$

p.55

— Solutions to K. Becker, M. Becker, J. Schwarz String Theory And M-theory

— Mikhail Goykhman

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What is the meaning of $\displaystyle{\alpha'}$?

How come $\displaystyle{\alpha' = \frac{1}{2}}$?

— Me@2018-12-07 10:43:10 PM

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2018.12.07 Friday (c) All rights reserved by ACHK

# Problem 14.5c2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

.

At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$.

Similarly, the spacetime fermions are obtained by tensoring all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = 4k}$ with the right-moving R- states with $\displaystyle{\alpha' M_R^2 = k}$.

c) Are there tachyonic states in heterotic string theory?

~~~

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— This answer is my guess. —

.

The left NS’+ sector:

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

.

The left R’+ sector:

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}

.

The right-moving NS+ states:

NS+ equations of (14.38):

\displaystyle{\begin{aligned} \alpha'M^2=0, ~~~&N^\perp = \frac{1}{2}: &b_{-1/2}^I~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\ \alpha'M^2=1, ~~~&N^\perp = \frac{3}{2}: &\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle, \\\ \alpha'M^2=2, ~~~&N^\perp = \frac{5}{2}: &\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}~& \\ &&\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \}~&|NS \rangle \otimes |p^+, \vec p_T \rangle \end{aligned}}

.

The R- states (that used as right-moving states):

Mass levels of R- and R+ (Equations 14.54):

\displaystyle{\begin{aligned} \alpha'M^2=0,~~~&N^\perp = 0:~~~~&|R_a \rangle~~&||~~|R_{\bar a} \rangle \\ \alpha'M^2=1,~~~&N^\perp = 1:~~~~&\alpha_{-1}^I |R_{a} \rangle,~d_{-1}^I |R_{\bar a} \rangle ~~&||~~ ... \\ \alpha'M^2=2,~~~&N^\perp = 2:~~~~&\{ \alpha_{-2}^I,~\alpha_{-1}^I \alpha_{-1}^J,~d^I_{-1} d^J_{-1} \} |R_{a} \rangle,~&|| \\ &&\{\alpha_{-1}^I d_{-1}^J,~d_{-2}^I \} |R_{\bar a} \rangle~~&||~~ ... \\ \end{aligned}}

.

There are no tachyonic states in heterotic string theory, since neither of the right-moving parts (NS+ and R-) has states with \displaystyle{\begin{aligned} \alpha' M^2 < 0\end{aligned}}.

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— This answer is my guess. —

— Me@2018-11-22 12:00:30 PM

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2018.11.22 Thursday (c) All rights reserved by ACHK

# Problem 14.5c

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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At any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string, the spacetime bosons are obtained by “tensoring” all the left states (NS’+ and R’+) with $\displaystyle{\alpha' M_L^2 = k}$ with the right-moving NS+ states with $\displaystyle{\alpha' M_R^2 = k}$.

~~~

In heterotic (closed) string theory, there are left-moving part and right-moving part. Then, what is the meaning of “at any mass level $\displaystyle{\alpha' M^2 = 4k}$ of the heterotic string”?

— Me@2018-11-11 03:44:18 PM

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Type IIA/B closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322

$\displaystyle{\alpha' M_L^2 = \alpha' M_R^2}$

$\displaystyle{\frac{1}{2} \alpha' M^2 = \alpha' M_L^2 + \alpha' M_R^2}$

$\displaystyle{\alpha' M^2 = 2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right) = 4 \alpha' M_L^2}$

These are the reasons that any mass level of the heterotic string is always in the form $\displaystyle{\alpha' M^2 = 4k}$.

— Me@2018-11-12 03:09:11 PM

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Equation (14.77):

p.322

closed string sectors: (NS, NS), (NS, R), (R, NS), (R, R)

$\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)$

$\text{type IIB}:~~~(NS+, NS+), ~(NS+, R-),~ (R-, NS+), ~ (R-, R-)$

What is the difference between Type IIA/B closed superstrings and heterotic SO(32) strings?

— Me@2018-11-12 03:15:46 PM

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The five consistent superstring theories are:

• The type I string has one supersymmetry in the ten-dimensional sense (16 supercharges). This theory is special in the sense that it is based on unoriented open and closed strings, while the rest are based on oriented closed strings.
• The type II string theories have two supersymmetries in the ten-dimensional sense (32 supercharges). There are actually two kinds of type II strings called type IIA and type IIB. They differ mainly in the fact that the IIA theory is non-chiral (parity conserving) while the IIB theory is chiral (parity violating).
• The heterotic string theories are based on a peculiar hybrid of a type I superstring and a bosonic string. There are two kinds of heterotic strings differing in their ten-dimensional gauge groups: the heterotic E8×E8 string and the heterotic SO(32) string. (The name heterotic SO(32) is slightly inaccurate since among the SO(32) Lie groups, string theory singles out a quotient Spin(32)/Z2 that is not equivalent to SO(32).)

— Wikipedia on Superstring theory

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2018.11.15 Thursday (c) All rights reserved by ACHK

# Problem 14.5b2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) … Keep only states with $\displaystyle{(-1)^{F_L}=+1}$; this defines the left R’+ sector.

Write explicitly and count the states we keep for the two lowest mass levels, indicating the corresponding values of $\displaystyle{\alpha' M_L^2}$. [This is a shorter list.]

~~~

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

If we define $N^\perp$ in a way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + N^\perp \\ \end{aligned}}

.

\displaystyle{\begin{aligned} (-1)^{F_L} |R_\alpha \rangle_L &= + |R_\alpha \rangle_L \\ (-1)^{F_L} |R_\alpha \rangle_R &= - |R_\alpha \rangle_L \\ \end{aligned}}

.

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 0:~~~~~&|R_\alpha \rangle_L \\ \alpha'M^2=2,~~~&N^\perp = 1:~~~~~&\alpha_{-1} |R_\alpha \rangle_L, \lambda_{-1} |R_\alpha \rangle_R \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-11-06 03:39:15 PM

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2018.11.06 Tuesday (c) All rights reserved by ACHK

# Problem 14.5b2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

We have 32 zero modes $\displaystyle{\lambda_0^A}$ and 16 linear combinations behave as creation operators.

As usual half of the ground states have $\displaystyle{(-1)^{F_L} = +1}$ and the other half have $\displaystyle{(-1)^{F_L} = -1}$.

Let $\displaystyle{|R_\alpha \rangle_L}$ denote ground states with $\displaystyle{(-1)^{F_L} = +1}$.

How many ground states $\displaystyle{|R_\alpha \rangle_L}$ are there?

~~~

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

p.315 “Being zero modes, these creation operators do not contribute to the mass-squared of the states. Postulating a unique vacuum $\displaystyle{|0 \rangle}$, the creation operators allow us to construct $\displaystyle{16 = 2^4}$ degenerate Ramond ground states.”

Following the same logic:

Postulating a unique vacuum $\displaystyle{|0 \rangle}$, the creation operators allow us to construct $\displaystyle{2^{16}}$ degenerate Ramond ground states.

Therefore, there are $\displaystyle{2^{15}}$ ground states $\displaystyle{|R_\alpha \rangle_L}$.

— This answer is my guess. —

— Me@2018-10-29 03:11:07 PM

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2018.10.29 Monday (c) All rights reserved by ACHK

# Problem 14.5b

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(b) Consider the left R’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

— This answer is my guess. —

The naive mass formula in the left R’ sector:

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = -1, -2, ...} n \lambda_{-n}^A \lambda_n^A + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\ \end{aligned}}

p.316 Equation (14.51):

\displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n = -1. -2, ...} n \lambda_{-n}^A \lambda_n^A &= \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} (D - 2) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A &= \left[ \frac{1}{2} \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{1}{24} \left[(D - 2)\right]_A \right] + \frac{1}{2} \sum_{n = 1, 2, ...} n \lambda_{-n}^A \lambda_n^A \\ &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{32}{24} \\ &= \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \\ \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{n \ne 0} n \lambda_{-n}^A \lambda_n^A \\ &= \left[ \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \left[ \sum_{n = 1. 2, ...} n \lambda_{-n}^A \lambda_{n}^A + \frac{4}{3} \right] \\ &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

.

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= 1 + \sum_{n \in \mathbf{Z}^+} \left( \bar \alpha_{-n}^I \bar \alpha_{n}^I + n \lambda_{-n}^A \lambda_{n}^A \right) \\ \end{aligned}}

— This answer is my guess. —

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2018.10.24 Wednesday (c) All rights reserved by ACHK

# Problem 14.5a4

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

The GSO projection here keeps the states with $\displaystyle{(-1)^{F_L} = + 1}$; this defines the left NS’+ sector.

Write explicitly and count the states we keep for the three lowest mass levels, indicating the corresponding values of $\displaystyle{\alpha' M_L^2}$. [This is a long list.]

~~~

p.314 “Let us declare that number to be minus one, thus making the ground states fermionic:”

Equation (14.39):

$\displaystyle{(-1)^F |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle = - |NS \rangle \otimes |p^+, \overrightarrow{p}_T \rangle}$

Equation (14.40):

$\displaystyle{(-1)^F |\lambda \rangle = -(-1)^{\sum_{r,J} \rho_{r,J}} |\lambda \rangle}$

p.315 “So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = -1}$; they are fermionic states.”

However, in this problem:

“The left NS’ sector is built with oscillators $\displaystyle{\bar \alpha_{-n}^I}$ and $\displaystyle{\lambda_{-r}^A}$ acting on the vacuum $\displaystyle{|NS' \rangle_L}$, declared to have $\displaystyle{(-1)^{F_L} = + 1}$:”

$\displaystyle{(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L}$

So all the states with integer $\displaystyle{N^{\perp}}$ have $\displaystyle{(-1)^F = +1}$.

— This answer is my guess. —

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

If we define $N^\perp$ in the way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + N^\perp \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&|NS' \rangle_L, \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&\{ \bar \alpha_{-1}^I , \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \}|NS' \rangle_L, \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\{ \bar \alpha_{-1}^I \bar \alpha_{-1}^J, \bar \alpha_{-1}^I \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B, \lambda_{\frac{-3}{2}}^A \lambda_{\frac{-1}{2}}^B, ... \} \\ & & \{ ..., \lambda_{\frac{-1}{2}}^A \lambda_{\frac{-1}{2}}^B \lambda_{\frac{-1}{2}}^C \lambda_{\frac{-1}{2}}^D \} |NS' \rangle_L \\ \end{aligned}}

Let $\displaystyle{N(n, k) = {n + k - 1 \choose k - 1}}$, the number of ways to put n indistinguishable balls into k boxes.

\displaystyle{\begin{aligned} \alpha'M^2=-1,~~~&N^\perp = 0:~~~~~&1 \\ \alpha'M^2=0,~~~&N^\perp = 1:~~~~~&8 + \frac{32 \times 31}{2} = 504 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&\left( \frac{8 \times 8}{2} + \frac{8}{2} \right) = 36, 8 \times \left( \frac{32 \times 31}{2} \right) = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \end{aligned}}

\displaystyle{\begin{aligned} \alpha'M^2=1,~~~&N^\perp = 2:~~~~~&N(2,8) = 36, 8 \times {32 \choose 2} = 3968, 32 \times 32 = 1024, {32 \choose 4} = 35960 \\ \alpha'M^2=1,~~~&N^\perp = 2:~~~~~& 36 + 3968 + 1024 + 35960 = 40988 \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-14 03:25:08 PM

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2018.10.16 Tuesday (c) All rights reserved by ACHK

# Problem 14.5a4

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

.

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

— This answer is my guess. —

.

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{24} \left[ (D - 2) \right]_{I= 2, 3, ..., 9} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

$\displaystyle{\alpha' M_L^2 = \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

Equation (14.34):

$\displaystyle{\frac{1}{2} \sum_{r=- \frac{1}{2}, -\frac{3}{2}} r b_{-r}^I b_{r}^I = \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r b_{-r}^I b_{r}^I - \frac{1}{48} (D - 2)}$

.

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \frac{1}{2} \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \left[ \frac{1}{2} \sum_{r=\frac{1}{2}, \frac{3}{2}} r \lambda_{-r}^A \lambda_{r}^A - \frac{1}{48} \left[D - 2\right]_A \right]}$

$\displaystyle{= \frac{-1}{3} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A - \frac{32}{48}}$

.

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= -1 + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

— This answer is my guess. —

— Me@2018-10-10 05:38:08 PM

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2018.10.10 Wednesday (c) All rights reserved by ACHK

# Problem 14.5a3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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Paraphrasing the description of heterotic (closed) string theory:

• right-moving part $\sim$ an open superstring theory
• $NS$ sector: $\alpha_{-r}^I, b_{-r}^I,~~~I = 2,3,...,9$
• $R$ sector: $\alpha_{-n}^I, d_{-n}^I,~~~I=2,3,...,9$
• “The standard GSO projection down to NS+ and R- applies.”

.

• left-moving part $\sim$ a peculiar bosonic openstring theory
• $I = 2, 3, ..., 23:$ There are totally 24 transverse coordinates
• 8 bosonic coordinates $X^I$ with oscillators $\bar \alpha_{-n}^I$
• 16 peculiar bosonic coordinates

• can be replaced by 32 two-dimensional left-moving fermion fields, $\lambda^A$
• $\lambda^A$ (anti-commuting) fermion fields $\to$ has $NS'$ and $R'$ sectors
• .

• $NS'$: oscillators $\bar \alpha_{-n}^I, \lambda_{-r}^A$ act on the vacuum $|NS' \rangle$

given $(-1)^{F_L} |NS' \rangle_L = + |NS' \rangle_L \to$ defines the left $NS'+$ sector

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbb{Z} + \frac{1}{2}} r \lambda_{-r}^A \lambda_r^A}$

• $R'$: oscillators $\bar \alpha_{-n}^I, \lambda_{-n}^A$ act on $R'$ ground states

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \left( \bar \alpha_{-n}^I \bar \alpha_n^I + n \lambda_{-n}^A \lambda_n^A \right)}$

• .

— Me@2018-09-20 09:51:17 PM

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2018.09.21 Friday (c) All rights reserved by ACHK

# Problem 14.5a2

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

.

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

— This answer is my guess. —

.

Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$
$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

.

Equation (13.116):

$\displaystyle{\sum_{k \in \mathbf{Z}^+_{\text{odd}}} k = \frac{1}{12}}$

.

\displaystyle{\begin{aligned} &\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A \\ &= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right] \\ &= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{2} \sum_{r = 1, 3, ...} r + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ &= - \frac{1}{24} + 2 \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \end{aligned}}

.

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A \\ \end{aligned}}

.

If we define $N^\perp$ in the way similar to equation (14.37), we have

\displaystyle{ \begin{aligned} \alpha' M_L^2 &= \frac{-7}{48} + N^\perp \\ \end{aligned}}

— This answer is my guess. —

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— Me@2018-09-01 06:05:29 AM

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2018.09.01 Saturday (c) All rights reserved by ACHK

# Problem 14.5a1

Counting states in heterotic $SO(32)$ string theory | A First Course in String Theory

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(a) Consider the left NS’ sector. Write the precise mass-squared formula with normal-ordered oscillators and the appropriate normal-ordering constant.

~~~

.

$\displaystyle{\alpha' M_L^2 = \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

What is normal-ordering?

Put all the creation operators on the left.

.

What for?

p.251 “It is useful to work with normal-ordered operators since they act in a simple manner on the vacuum state. We cannot use operators that do not have a well defined action on the vacuum state.”

“The vacuum expectation value of a normal ordered product of creation and annihilation operators is zero. This is because, denoting the vacuum state by $|0\rangle$, the creation and annihilation operators satisfy”

$\displaystyle{\langle 0 | \hat{a}^\dagger = 0 \qquad \textrm{and} \qquad \hat{a} |0\rangle = 0}$

— Wikipedia on Normal order

.

— This answer is my guess. —

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^-} \bar \alpha_{-n}^I \bar \alpha_n^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \bar \alpha_{n}^I \bar \alpha_{-n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I \bar \alpha_{-n}^I - \bar \alpha_{-n}^I \bar \alpha_{n}^I + \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

.

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} \left[ \bar \alpha_{n}^I, \bar \alpha_{-n}^I \right] + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$= \displaystyle{\sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

.

c.f. p.251:

$\displaystyle{\sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I}$

$\displaystyle{= \sum_{n \in \mathbf{Z}^+} n \eta^{II} + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

$\displaystyle{= \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I}$

.

Equation at Problem 14.5:

$\displaystyle{\alpha' M_L^2}$

$\displaystyle{= \frac{1}{2} \sum_{n \ne 0} \bar \alpha_{-n}^I \bar \alpha_n^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{1}{2} \left[ \frac{-1}{12} (D - 2) + 2 \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I \right] + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{24} (D - 2) + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \frac{-1}{8} + \sum_{n \in \mathbf{Z}^+} \bar \alpha_{-n}^I \bar \alpha_{n}^I + \frac{1}{2} \sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

.

$D = 10$

.

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = - \frac{1}{2}, - \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} (-r) \lambda_{r}^A \lambda_{-r}^A + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

.

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A, \lambda_r^A \right]}$

.

Equation (14.29):

$\displaystyle{\left\{ b_r^I, b_s^J \right\} = \delta_{r+s, 0} \delta^{IJ}}$

$\displaystyle{b_r^I b_s^J = - b_s^I b_r^J + \delta_{r+s, 0} \delta^{IJ}}$

.

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \lambda_{r}^A \lambda_{-r}^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ (-1) \left( - \lambda_{-r}^A \lambda_r^A + \delta_{r-r, 0} \delta^{AA} \right) + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ \lambda_{-r}^A \lambda_r^A + \lambda_{-r}^A \lambda_r^A - 1 \right]}$

$\displaystyle{= \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ 2 \lambda_{-r}^A \lambda_r^A - 1 \right]}$

.

$\displaystyle{\sum_{r \in \mathbf{Z} + \frac{1}{2}}r \lambda_{-r}^A \lambda_r^A}$

$\displaystyle{= - \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= - \frac{1}{2} \sum_{r = 1, 3, ...} r + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left[ b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right]}$

$\displaystyle{= \left[ - \frac{1}{24} + \sum_{r = \frac{1}{2}, \frac{3}{2}, ...} r \left( b_{-r}^A b_r^A + \lambda_{-r}^A \lambda_r^A \right) \right]}$

— This answer is my guess. —

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— Me@2018-08-06 10:23:48 PM

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2018.08.06 Monday (c) All rights reserved by ACHK

# Problem 14.4b2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

Type IIB closed superstrings

Equation (14.85)

$(NS+, NS+), (NS+, R-), (R-, NS+), (R-, R-)$

— Me@2015.09.16 06:08 AM: Should be the same. But I am not sure whether I have missed something.

.

$f_{NS+}(x) = 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...$

$f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

$f_{NS-}(x) = \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...$

$f_{R+}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

— Me@2018-07-14 09:41:10 PM

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2018.07.14 Saturday (c) All rights reserved by ACHK

# Quick Calculation 14.8.2

A First Course in String Theory

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What sector(s) can be combined with a left-moving NS- to form a consistent closed string sector?

~~~

There are no mass levels in NS+, R+, or R- that can match those in NS-. So NS- can be paired only with NS-:

$(NS-, NS-)$

.

$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= \frac{1}{\sqrt{x}} g_{NS}(x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

.

$g (\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

$g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

.

$g (\sqrt{x}) + g (-\sqrt{x})$
$= 2(1 + 36 x + 402 x^{2} + 3064 x^{3} + ...)$

.

$f_{NS-}(x)$
$= \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) + g (-\sqrt{x}) \right]$
$= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 + \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]$
$= \frac{1}{2 \sqrt{x}} \left[ 2(1 + 36 \, x + 402 \, x^{2} + 3064 \, x^{3} + ...) \right]$
$= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x^{\frac{3}{2}} + 3064 x^{\frac{5}{2}} + ...$

— Me@2018-06-26 07:36:41 PM

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2018.06.26 Tuesday (c) All rights reserved by ACHK

# Problem 14.4b1.4

Closed string degeneracies | A First Course in String Theory

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What is the meaning of “With $a = 1, ..., 8$ and $\bar b = \bar 1, ..., \bar 8$, …”?

— Me@2015.09.14 12:11 PM

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p.315 “Explicitly, the eight states $| R_a \rangle, a = 1, 2, ..., 8$, with an even number of creation operators are … ”

p.316 “The eight states $|R_{\bar{a}} \rangle, \bar a = \bar 1, \bar 2, ..., \bar 8$, with an odd number of creation operators are … ”

— Me@2018-05-24 11:41:34 AM

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2018.05.24 Thursday (c) All rights reserved by ACHK

# Problem 14.4b1.3

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

— This answer is my guess. —

Since for NS, the first 5 levels’ degeneracies are 8, 36, 128, 402, 1152, the degeneracies of (NS, NS) are $8^2, 36^2, 128^2, 402^2, 1152^2$.

This is incorrect, for there are no (NS, NS) states. Instead, you should consider (NS+, NS+).

.

Since for NS+, the first 5 levels’ degeneracies are 8, 128, 1152, 7680, 42112, the degeneracies of (NS+, NS+) are $8^2, 128^2, 1152^2, 7680^2, 42112^2$.

p.317 Consider the relationship of the degeneracy of R+ and that of R-:

.

How about the first 5 levels of R+?

The degeneracies are the same as those of R-.

p.317 Equation (14.54) “The appearance of an equal number of bosonic and fermionic states at every mass level is a signal of supersymmetry. This is, however, supersymmetry on the world-sheet.”

Equation (14.71):

$f_{R-}(x) = 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ...$

p.321 “Indeed, the integer mass-squared levels in the NS generating function (14.67) have degeneracies that match those of (14.71) for the R- sector.”

— This answer is my guess. —

— Me@2018-05-14 02:51:55 PM

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2018.05.14 Monday (c) All rights reserved by ACHK

# Problem 14.4b1.2

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

~~~

NS+ equations of (14.38):

 $\alpha'M^2=0,$ $~~N^\perp = \frac{1}{2}:$ $~~~~b_{-1/2}^I |NS \rangle \otimes |p^+, \vec p_T \rangle,$ $\alpha'M^2=1,$ $~~N^\perp = \frac{3}{2}:$ $~~~~\{ \alpha_{-1}^I b_{\frac{-1}{2}}^J, b_{\frac{-3}{2}}^I, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K \} |NS \rangle \otimes |p^+, \vec p_T \rangle,$ $\alpha'M^2=2,$ $~~N^\perp = \frac{5}{2}:$ $~~~~\{\alpha_{-2}^I b_{\frac{-1}{2}}^J, \alpha_{-1}^I \alpha_{-1}^J b_{\frac{-1}{2}}^K, \alpha_{-1}^I b_{\frac{-3}{2}}^J, \alpha_{-1}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M, ...\}$ $\{ ..., b_{\frac{-5}{2}}^I, b_{\frac{-3}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K, b_{\frac{-1}{2}}^I b_{\frac{-1}{2}}^J b_{\frac{-1}{2}}^K b_{\frac{-1}{2}}^M b_{\frac{-1}{2}}^N \} |NS \rangle \otimes |p^+, \vec p_T \rangle,$ … … …

For $N^\perp = \frac{5}{2}$, the number of states is

$8^2 + \left[ \frac{(8)(7)}{2!} + 8 \right] (8) + 8^2$
$+ 8 + \frac{(8)(7)(6)}{3!} + 8 + (8) \left[ \frac{(8)(7)}{2!} \right] + \frac{(8)(7)(6)(5)(4)}{5!}$
$= 1152$

.

Since $\alpha' M^2 = N^\perp - \frac{1}{2}$, when $N^\perp = \frac{5}{2}$, $\alpha' M^2 = 2$.

Equation (14.67):

$f_{NS} (x) = \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

Equation (14.66):

$f_{NS} (x) = \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

.

p.321

If we take $f_{NS} (x)$ in (14.66) and change the sign inside each factor in the numerator

Equation (14.72):

$\frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

the _only_ effect is changing the sign of each term in the generating function whose states arise with an odd number of fermions.

.

$\prod_{n=1}^\infty \left( \frac{1}{1-x^n} \right)^8$ is the boson contribution.

$\prod_{n=1}^\infty \left( 1+x^{n-\frac{1}{2}} \right)^8$ is the fermion contribution.

.

Turning Equation (14.67) into

$f_{NS?} (x)$
$= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...$

is equivalent to turning all $\sqrt{x}$ into $- \sqrt{x}$:

$f_{NS?} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$

Me@2015.08.29 12:49 PM: Somehow, $\sqrt{x}$ represents “contribution from fermions”.

.

Me@2015.08.29 12:50 PM: If you still cannot understand, try replace all $\sqrt{x}$ with $y$.

$f_{NS+} (x) = \frac{1}{2} \left( f_{NS} - f_{NS?} \right)$

$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

$f_{NS?} (x)$
$= - \frac{1}{\sqrt{x}} + 8 - 36 \sqrt{x} + 128 x - 402 x \sqrt{x} + 1152 x^2 - ...$

$f_{NS+} (x)$
$= \frac{1}{2} \left( f_{NS} - f_{NS?} \right)$
$= \frac{1}{\sqrt{x}} + 36 \sqrt{x} + 402 x \sqrt{x} + ...$

.

It is _not_ correct. Just consider it as $\left(\sqrt{x} \to -\sqrt{x} \right)$ is not correct, since the beginning factor $\frac{1}{\sqrt{x}}$ is not considered yet.

Instead, we should present in the following way:

$f_{NS} (x)$
$= \frac{1}{\sqrt{x}} \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= \frac{1}{\sqrt{x}} g_{NS}(x)$
$= \frac{1}{\sqrt{x}} + 8 + 36 \sqrt{x} + 128 x + 402 x \sqrt{x} + 1152 x^2 + ...$

$g (\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 + 8 \, \sqrt{x} + 36 \, x + 128 \, x^{\frac{3}{2}} + 402 \, x^{2} + 1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

$g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 1 -8 \, \sqrt{x} + 36 \, x -128 \, x^{\frac{3}{2}} + 402 \, x^{2} -1152 \, x^{\frac{5}{2}} + 3064 \, x^{3} + ...$

.

$g (\sqrt{x}) - g (-\sqrt{x})$
$= \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8$
$= 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ...$

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$f_{NS+}(x) = \frac{1}{2 \sqrt{x}} \left[ g (\sqrt{x}) - g (-\sqrt{x}) \right]$
$= \frac{1}{2 \sqrt{x}} \left[ \prod_{n=1}^\infty \left( \frac{1+x^{n-\frac{1}{2}}}{1-x^n} \right)^8 - \prod_{n=1}^\infty \left( \frac{1-x^{n-\frac{1}{2}}}{1-x^n} \right)^8 \right]$
$= \frac{1}{2 \sqrt{x}} \left[ 16 \, \sqrt{x} + 256 \, x^{\frac{3}{2}} + 2304 \, x^{\frac{5}{2}} + 15360 \, x^{\frac{7}{2}} + ... \right]$
$= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ...$

— Me@2018-05-08 08:50:32 PM

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2018.05.08 Tuesday (c) All rights reserved by ACHK

# Problem 14.4b1.1

Closed string degeneracies | A First Course in String Theory

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(b) State the values of $\alpha' M^2$ and give the separate degeneracies of bosons and fermions for the first five mass levels of the type IIA closed superstrings. Would the answer have the different for type IIB?

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Type IIA closed superstrings

p.322

In closed superstring theories spacetime bosons arise from the (NS, NS) sector and also from the (R, R) sector, since this sector is “doubly” fermionic. The spacetime fermions arise from the (NS, R) and (R, NS) sectors.

p.322 $\alpha' M_L^2 = \alpha' M_R^2$

 $\frac{1}{2} \alpha' M^2 =$ $\alpha' M_L^2 + \alpha' M_R^2$ $\alpha' M^2 =$ $2 \left( \alpha' M_L^2 + \alpha' M_R^2 \right)$ $=$ $4 \alpha' M_L^2$

Equation (14.77):

$\text{type IIA}:~~~(NS+, NS+), ~(NS+, R+),~ (R-, NS+), ~ (R-, R+)$

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What is the difference of the meanings of R+ and R-?

$R+$ states are world-sheet bosonic states.

p.316

It thus follows that all eight $| R_a \rangle$ states are fermionic and all $| R_{\bar a} \rangle$ are bosonic.

Be careful:

Here, “fermionic”/”bosonic” refers to the world-sheet fermions/bosons, not the spacetime ones.

p.320

Identifying $| R_a \rangle$ as spacetime fermions and $| R_{\bar a} \rangle$ as spacetime bosons is not an alternative either, since spacetime bosons cannot carry a spinor index.

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How come the R+ cannot be the left-moving part?

p.320

A strategy then emerges. Since all states in the R sector have a spinor index, we will only attempt to get spacetime fermions from this sector. We also recognize that all fermions must arise from states with the same value of $(-1)^F$.

Me@2015.09.11 10:36 AM: In other words, it is a convention:

Following Gliozzi, Scherk, and Olive (GSO) we proceed to truncate the Ramond sector down to the set of states with $(-1)^F = -1$.

— Me@2018-05-01 05:59:53 PM

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2018.05.01 Tuesday (c) All rights reserved by ACHK